Answer
$ \dfrac{- \pi }{4}$
Work Step by Step
We are given: $\sin^{-1} [\cos (\dfrac{ 3 \pi}{4})]$
We know that the trigonometric function $\sin^{-1} (1)$ has domain $[-1,1]$ and range: $[-\dfrac{\pi}{2}, \dfrac{\pi}{2}]$.
Here, $\sin^{-1} (\dfrac{-\sqrt 2}{2})=\dfrac{-\pi}{4}$, because $\sin{(\dfrac{-\pi}{4})}=\dfrac{-\sqrt 2}{2}$ and $\dfrac{-\pi}{4}$ lies in the range of $\sin^{-1}{x}$ .
Therefore, we have:
$\sin^{-1} [\cos (\dfrac{ 3 \pi}{4})]=\sin^{-1} ({\dfrac{-\sqrt 2}{2}})=\dfrac{- \pi }{4}$
So, $\theta = \dfrac{- \pi }{4}$
